I'm working through an openstax textbook "A First Course in Electrical and Computer Engineering" and stuck on two similar problems relating to multiplicative inverses in section 3.3
A) Show that $z^{-1}$ may be written as $r^{-1}e^{-j\theta}$
From my understanding:
$z^{-1} = \frac{x -jy}{x^2 + y^2}$ (as defined in the text)
$r = \sqrt{x^2 + y^2}$
$e^{-j\theta} = x - jy$
so:
$r^{-1}e^{-j\theta} = (\frac{1}{\sqrt{x^2 + y^2}}) * (x-jy) = \frac{x - jy}{\sqrt{x^2 + y^2}} \neq \frac{x -jy}{x^2 + y^2}$
B) Prove $(z^{-1})^* = (z^*)^{-1} = r^{-1}e^{j\theta} = \frac{1}{z^*z}z$
Just worrying about resolving the $r^{-1}e^{j\theta}$ term here and comparing it to the easier to solve $\frac{1}{z^*z}z$:
$\frac{1}{z^*z}z = \frac{x + jy}{(x-jy)(x+jy)} = \frac{x + jy}{x^2 + y^2}$
$r^{-1}e^{j\theta} = (\frac{1}{\sqrt{x^2 + y^2}})(x + jy) = \frac{x+jy}{\sqrt{x^2 + y^2}} \neq \frac{x + jy}{x^2 + y^2}$
I'm guessing I must be making a mistake with converting the $r^{-1}e^{j\theta}$ term into cartesian representation, but I've been stuck for a while without seeing the solution.
Edit: It seems my issue is that I was using an incorrect formula for $e^{j\theta} = cos \theta + j sin\theta = x + jy$ which would only be valid if $r=1$
Correctly using the formula $e^{j\theta} = r cos \theta + j r sin \theta = \frac{x}{r} + j\frac{y}{r} = \frac{x + jy}{\sqrt{x^2 + y^2}}$ produces the right answers to both questions.